1995 IMO Problem #1

[devendraneupane]

Let's call the concurrent point L. Now, Instead of using the radical axis theorem, can you please tell me if the alternative process is correct?
Angle AMC is right angle and so is angle BND, so point P is the orthocenter of Triangle LAD. Since XY is perpendicular to AD, XZ is the third altitude passing through the orthocenter P which must meet at the vertex of the triangle LAD, which is the L. Hence they are all concurrent.